Question

The equation of a sound wave in air is given by $\Delta p=0.09\sin (400t-9x)$ where all variable are in SI units. If the equilibrium pressure of air is $1$ atm, then

[$1\text{atm}=1.013\times {10}^5{\text{N/m}}^2$]

• A. Frequency of sound wave is $\frac{200}{\pi }\text{Hz}$.
• B. Speed of the sound wave is $90\text{m/s}$.
• C. Maximum pressure at a point is $(1.013\times {10}^5+0.09){\text{N/m}}^2$.
• D. Minimum pressure at a point is $(1.013\times {10}^5-0.09){\text{N/m}}^2$.

Answer 1

Answer: (A), (C), (D)

Minimum pressure at a point is $(1.013\times {10}^5-0.09){\text{N/m}}^2$.

Given equation is
$\Delta p=0.09\sin (400t-9x)$

Comparing with the standard form of travelling sound wave,

$\Delta p=\left(\Delta p{)}_{max}\sin \right(\omega t-kx)$

$(\Delta p_0{)}_{max}=0.09{\text{N/m}}^2$ , $\omega =400\text{rad/s}$ and $k=9{\text{m}}^{-1}$

So the frequency $\left(f\right)$ of the wave is

$f=\frac{\omega }{2\pi }=\frac{400}{2\pi }=\frac{200}{\pi }\text{Hz}$

Speed of sound $\left(v\right)$ is

$v=\frac{\omega }{k}=\frac{400}{9}=44.5\text{m/s}$

Since given that
Equilibrium pressure , $p_0=1\text{atm}=1.013\times {10}^5{\text{N/m}}^2$

Using , $(\Delta p_0{)}_{max}=0.09$

we get,

Maximum pressure $=(1.013\times {10}^5+0.09){\text{N/m}}^2)$

Minimum pressure $=(1.013\times {10}^5-0.09){\text{N/m}}^2)$

Thus, options (a), (c) and (d) are the correct answers.